A sealed vertical cylinder of radius R and height h = 0.567 m is initially fille

Question

A sealed vertical cylinder of radius R and height h = 0.567 m is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, p0 = 1.01·105 Pa. A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.)

Explanation / Answer

initial depth of water=0.567/2=0.2835 m

let the new depth of the water is h.

air pressure at the time when the cylinder stops leaking is P.

then pressure at the end of the water level=P+density of water*g*depth

=P+1000*9.8*(0.2835-h)=P+2778.3-9800*h....(1)

this pressure value is equal to atmospheric pressure as the cylinder is open at the bottom

hence P+2778.3-9800*h=1.01*10^5

==>P=98221.7+9800*h...(1)

let crosssectional area of the cylinder be A.

for the air portion, using ideal gas law,

P1*V1=P2*V2

where P1=initial pressure=1.01*10^5 Pa

V1=initial volume=Area*height=A*0.567/2=0.2835*A

P2=final pressure=P

V2=final volume=area*depth=A*(0.2835+h)

hence 1.01*10^5*A*0.2835=P*A*(0.2835+h)

==>28633.5=(98221.7+9800*h)*(0.2835+h)

==>28633.5=27845.85+1.01*10^5*h-9800*h^2

==>9800*h^2+1.01*10^5*h-787.65=0

solvign for h, we get

h=7.792 mm

==>h=

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.